How to prove that $ q^{(n)} = q^{(0)}P^{(n)}$ for Markov Chains This is where $P = \begin{bmatrix}
p_{11}&p_{12}\\
P_{21}&p_{22}\\
\end{bmatrix}$ , and $q_{i}^{(n)}=P(X_n=i)$
We derived this in lecture for my Stochastic Processes course, but unfortunately I missed that day and am finding the derivation surprisingly hard to find online. This is not a homework problem because it was derived in class, but I think I would like to understand it to make sure I have a solid foundation for further derivations with regards to Markov Chains. Thanks!
 A: By definition: the next iteration of the vector is the matrix product of current iteration of the vector and the transition matrix.
$$q^{(n)} = q^{(n-1)} P\\ \begin{bmatrix}q_1^{(n)} & q_2^{(n)}\end{bmatrix} = \begin{bmatrix}q_1^{(n-1)}&q_2^{(n-1)}\end{bmatrix}\begin{bmatrix}p_{11} & p_{12}\\ p_{21} & p_{22}\end{bmatrix}$$
Then since $P^0$ is the identity matrix: $${q^{(0)}=q^{(0)}P^0\\ q^{(1)}= q^{(0)}P\\ q^{(2)}= (q^{(0)}P)P\\\vdots\\q^{(n)}{{}=(\cdots((q^{(0)}P)P)\cdots P)_{n\text{ iterations}}\\{}= q^{(0)}P^n}}$$
$\Box$

More formaly, we have that $q^{(0)}=q^{(0)}P^0$ and if for all positive natural $n$ we can assume $q^{(n-1)}=q^{(0)}P^{n-1}$, then we find from the definition that: $$q^{(n)}~{=q^{(0)}P^{n-1}P\\= q^{(0)}P^n}$$
So therefore by weak induction:$$\left\lvert{\begin{align}&q^{(0)}=q^{(0)}P^0\\ &\forall n\in \Bbb N^+ : \big[q^{(n-1)}=q^{(0)}P^{n-1}~\to~ q^{(n)}=q^{(0)}P^n\big]\\\hline & \forall n\in \Bbb N: q^{(n)}=q^{(0)}P^n \end{align}}\right.$$
$\blacksquare$

Note: $q^{(n)}$ is not $q^n$.   The bracketed superscript indicates the $n$-th iteration, not the $n$-th exponentiation.
